Prepare an ensemble of identically prepared photon pair states, measure thrm to get interference information. Group all of them together to get the hologram formed by the wavefunction of a single photon

[Something else about inteepretarions] We define the wavefunction as something that contains all the information we can learn about the quantum state. After reading the PSE answers about probability in quantum

mechanics, I am tempted to think that a hermitian operator $\hat{M }$ kind of filter out the probability amplitude that corresponds to the information of the observable $m$ only, which then, when the overlap of that with the wavefunction is computed via the inner product $\langle \psi \lvert \hat{M} \rvert \psi \rangle$

with give the sum of all values of $m$ weighted by the amplitude squared (i.e probability), thus giving the expectation value as expected. However there is an obvious question to ask. We knew that

while there are a lot of observables are noncommutative, there are some that are. This means there exists some state $\lvert K \rangle such that it is both an eigenvector of two observables $\hat{A}, \hat{B}$. Then

the interpretation of hermitian operators "filtering" out relevant information from the wavefunction will be challenging in this special case: Suppose I have the ket $\lvert K \rangle$, does it mean that it contains information of both $a$ and $b$...? In that case we cannot really say a hermitian operator $\hat{A}$ filter out information of $a$ only from the wavefunction, as the above ket also contains information of $b$

Perhaps one way to look at it is similar to the mathematical phenomenon of some equation of motion having multiple solutions at some positions in phase space, it might just happened that under a certain basis, the state vector that corresponds to some observable $a$ happened to have the same form as $b$

And since it is true for one basis, it is true for all bases as eigenvectors of linear operators are independent of basis. So maybe, if we insist on the filtering interpretation, there are actually two $\lvert K\rangle$ that look the same but contains diffetent information of the wavefunction that corresponds to the quantum state it started with. An

alternate way to think about this issue is that most filters are not perfect (because they are linear hermitian operators, there are only so many possible states they can act on and not give the same state vector), therefore when a hermitian operator $\hat{A}$ is trying to filter information of $a$ it is more common that the filtered result will contain information of, say, $b$. This will then explain noncommutativity

Because any state vector is linear, they can be linear combinations of other state vectors, which may contains eigenvectors corresponds to some hermitian operator. Then assuming we have the filtering information interpeetation above, we will need to deal with the issue of how can information of $a$ (say position) be added to information of $b$ (say spin projected at the z axis) to give information of a completely unrelated $c$ (say the mass of a particle)

If the above interpretation is the correct one (which I doubt it) it will imply information has a weird property that they are linear, they give different physical qualities when arranged in certain ways

and then, one may come to the conclusion that all the physical things like charge, momentum etc might be nothing but a result of different ways these information overlap with each other. And in the macroscopic scale, decoherence (may be one of thr things that) ensure we only ever observed one of the many, with some probability

Unless there's an experimental way to probe this interpretarion about information, I don't think the filtering interpretation I had make sense

Therefore, either information has such weird properties, or that the filtering interpretation is wrong, thus meaning that quantum probabilities and logic cannot simply be treated in the same way (with some modification such as linearity and superposition) of claasical probability in the notion of that propositions filter out relevant elements from the set (except that by definition it just happened that a vector space is a type of set)

Or put it more directly, quantum probabilities cannot be thought of as some form of venn diagram

Also if spin basically describe the topology of SO(n) and in 3+1 spacetime, lorentz invariance result in spacelike separated field operators to tranform in a symmetric or antisymmetric way, then how is one of the two classes of action of SO(3) being assigned to what field operator.

For example, why do field operators that creates something with lepton number 1, charge -1, $mass m_e$ (i.e. Electron) will have a SO(3) topology that is of the class that corresponds to "rotations by 4$\pi$"...?

I might get my answer when I revise how the value of spin of particles are calculated, such as the Higgs, the photon etc.

Many string theorists would like to know more algebraic geometry. There are a few of us who know algebraic geometry at a pretty high level (not me) but many more who would like to learn more and feel it would help with their research but find the literature very difficult. I think the optimal sol...

user116211

I think the optimal solution would be to find such a string theorist and agree that you will teach them algebraic geometry if they will teach you string theory.

Wrong!
Here is Bourbaki document on algebraic geometry, taken from the now available Master's Archives: click on Autres rédactions, then on Chap.I Théorie globale élémentaire (91 p.)
This preliminary draft was apparently written (according to a penciled annotation on the first page) for Bourb...

MAFIA36790 thought it was inappropriate for a person (myself) with PTSD to ask people on the internet not to be abusive to them (as I did in the original post and he/she removed) so I abandoned this question immediately since honestly I think I ought to be able to ask people to not be abusive to ...

user116211

Weirder fact is that it got a upvote too ;P

user116211

I've left a comment though:

user116211

I really don't know why you are doing all these. I removed an unnecessary part of your question; we have Be nice policy on SE; so you don't have to explicitly mention that. I'm not bullying you in any way. The comment was generated from the review as it didn't answer the original question. Also, if you have any grievance, you can post it Meta Physics SE. — MAFIA3679021 mins ago

I have passed differential equations some years ago and I have forgotten that completely. I am not sure, but maybe this help that differentiate both side of the equation with respect to the t for getting rid of power of 2.

Through substituting for values of $\rho$ and $k$, I have:
$H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G C}{3a^4} + \frac{\Lambda c^2}{3}$
$a=a(t)$, and $a(t=0)=0$. Note that $C$ is a constant of integration, introduced from the substitution to eliminate $\rho$ - it is not the same as $c$. I'm not re...

also, it's funny that people immediately turn to john rennie for cosmology. He is the top user in the tag, but I'd wager he's top in every tag because he answer probably 33% of all questions asked. What people should do is turn to the top users with high score/answer ratios

highest would appear to be ted bunn, then pulsar. Lubos doesn't count because his presence on chat and the site in general is too unreliable

I've noticed that $\mathrm{GL}_n(\mathbb R)$ is not a connected space, because if it were $\det(\mathrm{GL}_n(\mathbb R))$ (where $\det$ is the function ascribing to each $n\times n$ matrix its determinant) would be a connected space too, since $\det$ is a continuous function. But $\det(\mathrm{G...

@0celo7 Ah, you want to know math. Ask someone else. I'm a cosmologist that was formerly an engineer. I know the cosmology concepts very well and how to perform the math I might be called on to use, but I never studied the pure mathematics. Math terminology and identities get me every time